# TAMU CHEN 304 - L11 (15 pages)

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## L11

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- Pages:
- 15
- School:
- Texas A&M University
- Course:
- Chen 304 - Chem Engr Fluid Ops

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CBE341 Lecture 11 10 6 Constitutive equations for a flowing fluid What have we derived thus far in the course Till now I focused on deriving conservation equations by using the principle of mass conservation and Newton s law of linear momentum balance Let us have a recap of the equations I first formulated the integral mass balance equation See equation 36 of my notes Then I deduced the differential form of this balance commonly referred to as the continuity equation See equation 43 for a general case For the special case of an incompressible fluid we obtain equation 46 Then I started with Newton s law for linear momentum balance I derived the integral form of this balance applicable to an arbitrary control volume See equation 56 of my notes Then I introduced the notion of a deviatoric stress tensor see equations 117 and 118 Finally I converted equation 56 to a differential form referred to as the Cauchy s equations of motion see equations 129 and 136 We will now discuss the additional information needed before we can use the continuity equation and the Cauchy s momentum balance equations to find detailed solutions for the velocity profiles in some flow problems The continuity equation and the Cauchy s momentum balance equations are reproduced below for convenience r rv 0 t 43 rv rvv rg P t t 87 129 Or equivalently r v rv v rg P t t 136 Many flow problems involve only modest pressure drops and the volume change is small even for gases Hence even gas flow problems can be treated as incompressible fluid flow problems with little loss of accuracy For incompressible fluids equation 43 simplifies to v 0 46 Aside If we consider examples such as flow of a gas through a very long pipe with correspondingly large pressure drops and appreciable volume changes then we retain equation 43 Even in these problems the effect of compressibility in the momentum balance equations namely 129 or 136 turns out to be weak provided we are not talking about flows involving high Mach numbers close to unity or larger In chemical engineering we rarely deal with such problems and so we can ignore the effect of compressibility in equations 129 and 136 In this course where we want to illustrate the use of balance equations to solve flow problems we will only consider essentially incompressible flows So we will discuss the constitutive equations for the stress tensor only for the case of incompressible fluids It was already mentioned that the viscous stress tensor is symmetric Note that the viscous stress depends on the velocity gradient For example consider the so called plane Couette flow 88 According to our notations the viscous traction exerted by the upper plate on the fluid sandwiched between the two plates is given by t n t where n is the unit normal pointing from the entity on which the traction is being exerted in this case the fluid towards the agent exerting the traction in this case the upper plate That is Viscous traction t n t e y t t yx e x t yye y t yz e z One can expect intuitively that in order to sustain this motion the upper plate must exert a force in the positive x direction on the fluid This force is simply the product of the area of contact between the fluid and the upper plate and t yx e x of the viscous traction This component of the shear stress t yx depends on the velocity gradient For Newtonian fluids t yx depends on both dv x dy dv y dv x and see equation 156 at the end dy dx of these notes however in this example vy 0 Based on this one can anticipate that in a more general flow problem where the flow may not be one dimensional the stress tensor will depend on all possible velocity gradients dvx dvx dv x dv y dv y dv y dvz dvz dvz This is equivalent to saying that the dx dy dz dx dy dz dx dy dz stress tensor depends on the velocity gradient t t v where v vi x i 1 3 j 1 3 89 j ei e j In general however the velocity gradient v is not symmetric For example in the plane Couette flow problem 8 of the nine components of the velocity gradient v are zero and dv x is the only non zero element However the stress tensor has to be dy symmetric This dilemma is resolved mathematically by demanding that the viscous stress tensor depends on the symmetric part of the velocity gradient tensor Recall that the symmetric part of a two dimensional square matrix A aij is simply aij aji 2 In the same manner the symmetric part of the velocity gradient tensor is simply 1 vi v j 1 1 T v symmetric eie j v v x i 2 2 i 1 3 2 x j j 1 3 140 where the superscript T denotes the transpose What is the physical meaning of this requirement Consider a fluid element as shown on the left below in figure a Let us suppose that after a little time the fluid element has translated and also changed the shape to assume the form shown on the right side Translation of a fluid element does not produce any stress absence of deformation The change of shape clearly implies that the various corners of the original fluid element have moved at different rates i e the velocities of the various corners of the fluid element are different That is there are some non zero entries in the velocity gradient tensor Now the change in the configuration of the fluid element where I am not interested in the translation any more consists of two parts a deformation and a rotation 90 Figure a This is illustrated below where we first deform the shape see figure b Figure b and subsequently rotate the element to get the final configuration see figure c 91 Figure c Physically the rotation of a fluid element does not produce any stress So the operation in figure c should not play any role in computing the viscous stress and only the first step matters This is precisely what is achieved by taking the symmetric part of the velocity gradient tensor Figure a shows the full effect of the velocity gradient tensor Figure b shows only the contribution of the symmetric part captured in equation 140 Figure c shows the effect of the remainder of the velocity gradient tensor which is to purely rotate the fluid element Constitutive Equations for an incompressible Newtonian Fluid t v v T 141 Introducing equation 141 into Cauchy s equation 136 we get r v rv v rg P v v T t 142 For constant r and the final term on the right hand side of equation 142 can be written as 2 v v 2 v as v 0 because the fluid has been assumed to be incompressible Thus we have r v rv v rg P 2 v t …

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